Question 30148: Question:
Solve the system: x^2+y^2=4
x+y=2
Possible solutions:
(a) [{2,-2), (-2,-2)]
(b) [{0,-2), (-2,0)]
(c) [(0,2), (2,0)]
(d) [(0,0), (2,-2)]
Please help thanks! Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! x^2+y^2=4
x+y=2
(c) looks like the answer
x + y = 2
y = 2 - x
x^2 + (2 - x)^2 = 4
x^2 + 4 -4x + x^2 = 4
2x^2 -4x = 0
x^2 -2x = 0
x(x - 2) = 0
this is true when x = 0 or x = 2
when x = 0,
x + y = 2
0 + y = 2
y = 2
(0, 2) is a solution
when x = 2,
x + y = 2
2 + y = 2
y = 0
(2, 0) is a solution
answer is (c)
